# Download e-book for kindle: On the stability of properly-degenerate hamiltonian systems by Biasco L., Chierchia L.

By Biasco L., Chierchia L.

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Additional resources for On the stability of properly-degenerate hamiltonian systems with two degrees of freedom

Example text

In such a case38 χ+ (E1 , (−π, π)) = (−π, π), χ+ (E1 , (0, ±i∞)) = (0, ±is+ (E1 )), χ+ (E1 , (±π, ±π ± iψ0 (E1 ))) = (±π, ±π ± is+ (E1 )), where s+ (E1 ) := √ 2( dP + −1 ) dE ∞ dψ E1 + ε(1 + cosh ψ) 0 . + In fact, it is χ+ (E1 , M+ p (E1 )) = Ts+ (E1 ) . Let, now, E = E1 + iE2 ∈ E . In this case, χ+ (E, Tσ ∩ M+ p (E)) ⊇ Ts( E,σ) where s+ (E, σ) := min Imχ+ (E, t + iσ) , t∈(−π,π) and, as in the case E = E1 , s+ (E1 , σ) = Imχ+ (E1 , iσ) := √ 2( dP + −1 ) dE σ dψ E1 + ε(1 + cosh ψ) 0 . It is easy to see that σ 0 σ dψ E1 + ε(1 + cosh ψ) ∼ 0 dψ ˜1 + εψ 2 E 1 ∼ √ ln 1 + ε ε σ ˜ E1 .

If Qn ∈ W1 (E0 ), then, continuity in q0√comes from the definition of χ− . Also, since pn ∈ C+ √ and Img(E0 , qn ) < 0, we have pn := 2 g(E0 , qn ) → 2 g(E0 , q0 ) =: p0 . Let us turn now to the more delicate case Qn ∈ W2 (E0 ). We shall consider the real case, E0 ∈ R, since the complex case is analogous but more clumsy because of the loss of symmetries. In this case ReQ0 = ±π/2 and Imq0 = 0. The points qn are such that Qn = π − χ− (E0 , qn ) and, by the ¯ n ∈ W1 (E0 ). Since Q ¯n → Q ¯ 0 = π − Q0 , symmetry of χ− , we have that43 χ− (E0 , q¯n ) = π − Q − −1 −1 ¯ ¯ one has that χ (E0 , q¯n ) → Q0 .

21), we find ln 1 + ε ˜1 σ E ln 1 + ε |E1 | − s (E1 , σ) ∼ , which implies that there exists a constant c such that, for any E1 with 2ε+η/2 < E1 < −η/2, one has σ . s− (E1 , σ) ≥ 2c ln(ε/η) In the general case, by the estimates on P − and its derivatives, one finds that, if E = E1 + iE2 ∈ E − , then 1 σ s− (E, σ) ≥ s− (E1 , σ) ≥ c =: s . 2 ln(ε/η) The lemma is proved also in the negative energy case. References [1] Arnold V. I. (editor): Encyclopaedia of Mathematical Sciences, Dynamical Systems III, Springer–Verlag 3 (1988) 32 [2] Arnold V.